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Computing the Fourier transform of 1/(1+x^2)^2 Watch

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    I want to compute the Fourier transform of 1/(1+x^2)^2 in \mathbb{R}. To this end, let f(x)=1/(1+x^2)^2. Then

    \displaystyle\hat{f}(\xi)=\int_{-\infty}^{\infty}\frac{1}{(1+x^{2  })^{2}}e^{ix\cdot\xi}\,dx.

    Normally one would proceed via integration by parts, but it seems to be a dead end no matter how you go about it. For instance, we can compute:

    \displaystyle\hat{f}(\xi)=\frac{  1}{2}e^{ix\cdot\xi}\left(\frac{x  }{x^2+1}+\arctan x\right)\bigg|_{x=-\infty}^{\infty}+\frac{i\xi}{2} \int_{-\infty}^{\infty}\left(\frac{x}{x  ^2+1}+\arctan x\right)\,dx

    Does anyone have a suggestion? Certainly, one can compute that \int_{-\infty}^{\infty}1/(1+x^2)^2\,dx=\pi/2, although I doubt this can be of any help.
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    (Original post by RamocitoMorales)
    I want to compute the Fourier transform of 1/(1+x^2)^2 in \mathbb{R}.
    If you can use standard results, then \mathcal{F}(\frac{1}{x^2+b^2})= \frac{\pi e^{-b\alpha}}{b} then use DUTIS on b.

    I haven't actually tried this myself but I think that'll work.
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    (Original post by atsruser)
    If you can use standard results, then \mathcal{F}(\frac{1}{x^2+b^2})= \frac{\pi e^{-b\alpha}}{b} then use DUTIS on b.

    I haven't actually tried this myself but I think that'll work.
    If you have the FT for 1/(1+x^2) you should also be able to finish using standard rules for the FT of a derivative and the FT of xf(x).
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    (Original post by atsruser)
    If you can use standard results, then \mathcal{F}(\frac{1}{x^2+b^2})= \frac{\pi e^{-b\alpha}}{b} then use DUTIS on b.

    I haven't actually tried this myself but I think that'll work.
    What is "DUTIS"?
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    (Original post by RamocitoMorales)
    What is "DUTIS"?
    Differentiation under the integral sign.
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    (Original post by RamocitoMorales)
    I want to compute the Fourier transform of 1/(1+x^2)^2 in \mathbb{R}. To this end, let f(x)=1/(1+x^2)^2. Then

    \displaystyle\hat{f}(\xi)=\int_{-\infty}^{\infty}\frac{1}{(1+x^{2  })^{2}}e^{ix\cdot\xi}\,dx.
    As a couple of alternatives:

    1. You can try a contour integral approach - not sure how tricky that is in this case though. However in this post:

    https://www.thestudentroom.co.uk/sho...&postcount=313

    TeeEm computes a similar but simpler integral via contour techniques - you could adapt this slightly to find the integral I suggested earlier, then finish via DUTIS.

    2. You can note that since your function is even, the FT reduces to 2\int_0^\infty \frac{\cos x}{(1+x^2)^2} \ dx then you can follow the working in this post:

    https://www.thestudentroom.co.uk/sho...&postcount=107

    where Kummer computes precisely that integral via a Laplace transform approach.
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    (Original post by DFranklin)
    If you have the FT for 1/(1+x^2) you should also be able to finish using standard rules for the FT of a derivative and the FT of xf(x).
    I can't quite see how you get it using those rules - I must be missing some obvious step.
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    (Original post by atsruser)
    I can't quite see how you get it using those rules - I must be missing some obvious step.
    Diffing gives you the FT of x/(1+x^2)^2, and then you can relate the FT of x/(1+x^2)^2 to the FT of 1(1+x^2)^2 using the "FT of xf(x)" rule.

    (I might be getting it wrong, of course, it's almost 30 years since I did this in earnest...)
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    (Original post by atsruser)
    2. You can note that since your function is even, the FT reduces to 2\int_0^\infty \frac{\cos x}{(1+x^2)^2} \ dx .
    Since the Fourier transform is \hat{f}(\xi), shouldn't there be a \xi there somewhere?
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    (Original post by RamocitoMorales)
    Since the Fourier transform is \hat{f}(\xi), shouldn't there be a \xi there somewhere?
    Yes, it should be \cos \xi x if \xi is your parameter.
 
 
 
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